InterviewSolution
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InterviewSolution
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If `y^x+x^y+x^x=a^b`, find `(dy)/(dx)`. |
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Answer» Let `u=x^(x),v=x^(y) and w=y^(x)`. Then, `y+v+w=a^(b)` `rArr(du)/(dx)+(dv)/(dx)+(dw)/(dx)=0." …(i) "[because a^(b)=" constant"]` Now, `u=x^(x)` `rArr logu=xlog x` `rArr(1)/(u).(du)/(dx)=x.(d)/(dx)(logx)+logx.(d)/(dx)(x)` `" [on differentiating both sides w.r.t. x]"` `rArr(du)/(dx)=u.{x.(1)/(x)+(logx).1}` `rArr(du)/(dx)=x^(x)(1+logx)." ...(ii)"` And, `v=x^(y)` `rArr log v=ylogx` `rArr(1)/(v).(dv)/(dx)=y.(d)/(dx)(logx)+(logx).(d)/(dx)(y)` `" [on differentiating both sides w.r.t. x]"` `rArr(dv)/(dx)=v.[y.(1)/(x)+(logx).(dy)/(dx)]` `rArr(dv)/(dx)=x^(y){(y)/(x)+(logx)(dy)/(dx)}." ...(iii)"` And, `w=y^(x)` `rArr logw=xlogy` `rArr(1)/(w).(dw)/(dx)=x.(d)/(dx)(logy)+(logy).(d)/(dx)(x)` `" [on differentiating both sides w.r.t. x]"` `rArr(dw)/(dx)=w.{x.(1)/(y).(dy)/(dx)+(logy).1}.` `rArr(dw)/(dx)=y^(x){(x)/(y).(dy)/(dx)+(logy)}." ...(iv)"` Using (ii), (iii) and (iv) in (i), we get `x^(x)(1+logx)+x^(y){(y)/(x)+(logx)(dy)/(dx)}+y^(x).{(x)/(y).(dy)/(dx)+(logy)}=0` `rArr{x^(x)(1+logx)+y.x^((y-1))+y^(x)(logy)}+{x^(y)(logx)+xy^((x-1))}(dy)/(dx)=0` `rArr(dy)/(dx)=(-{x^(x)(1+logx)+y.x^((y-1))+y^(x)(logy)})/({x^(y)(logx)+xy^((x-1))}).` |
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